Physics Projectile Motion

Question

Question: A landscape architect is planning an artificial waterfall in a city park. Water flowing at 1.75 m/s will leave the end of a horizontal channel at the top of a vertical wall h = 3.35 m high, and from there the water falls into a pool (see figure).

(a) Will the space behind the waterfall be wide enough for a pedestrian walkway? (Assume that the average pedestrian walkway is 1 m wide.)

 

so.... I solved for t=1.02. x=1.785.

 

If I adjust X and Y to scale, I get x=.112 y=.209. 

 

How do I solve for V?

 

Answer 1

y(t) = (-1/2)gt² + v_0yt + y₀

 

g = 9/8 m/s²

 

v_0y = 0 (vertical velocity is 0 at t=0)

 

v_0x = 1.75 m/s = constant since there is no horizontal acceleration

 

y₀ = 3.35 m

 

y(t) = -4.9t² + 3.35

 

At ground level, y = 0

 

0 = -4.9t² + 3.35 ⇒ t = √(3.35/4.9) s = 0.8268 s

 

x_max = v_0xt = (1.75 m/s)(0.8268 s) ≅ 1.45 m

 

It looks like you have enough horizontal room.